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Show that:Highpass = Original-Lowpass
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written 8.4 years ago by
teamques10
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When we apply the LPF on the image, the center pixel z5 changes to
$1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9]$
Original- Low pass $=z_5 - 1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9] \\ = z_5 - 1/9[z_1+z_2+z_3+z_4+z_5+z_6+z_7+z_8+z_9] \\ =8/9 z_5-1/9z_1-1/9z_2-1/9z_3-1/9z_4-1/9z_6-1/9z_7-1/9z_8-1/9z_9 \\ =1/9Χ \ high \ pass \ mask$
| -1 | -1 | -1 |
|---|---|---|
| -1 | 8 | -1 |
| -1 | -1 | -1 |
This is nothing but a high pass mask - • HPF alternates LOW frequency components and allows to pass High frequency components of the image.HPF is the sharpening Second order derivative filter. HPF image can be obtained by subtracting LPF image from original image.
Therefore, Original image = LPF image + HPF image
HPF image = Original image - LPF image
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